Three-Dimensional Virial Theorem (Quantum Mechanics)

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SUMMARY

The three-dimensional virial theorem in quantum mechanics is expressed as 2<t> = <r·∇V> for stationary states. The proof involves the operator Q = r·p, where the time derivative of the expectation value is zero for stationary states. By applying the commutation relations between the Hamiltonian and the position-momentum operator, the relationship is established through the manipulation of terms involving kinetic and potential energy operators. This theorem is crucial for understanding the dynamics of quantum systems.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically stationary states.
  • Familiarity with operator algebra in quantum mechanics.
  • Knowledge of commutation relations and their implications in quantum systems.
  • Proficiency in manipulating expectation values and their time derivatives.
NEXT STEPS
  • Study the derivation of the time-dependent Schrödinger equation.
  • Learn about the implications of the virial theorem in quantum mechanics.
  • Explore the role of commutation relations in quantum mechanics.
  • Investigate applications of the virial theorem in various quantum systems.
USEFUL FOR

Students of quantum mechanics, physicists focusing on theoretical physics, and researchers analyzing quantum systems and their dynamics will benefit from this discussion.

NeoDevin
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Homework Statement


(a) Prove the three-dimensional virial theorem:

[tex]2<t> = <r\cdot \nabla V>[/tex]

(for stationary states)

Homework Equations



Eq. 3.71 (not sure if this applies to 3 dimensions, but I think so)

\frac{d}{dt}<Q> = \frac{i}{\hbar}<[\hat H, \hat Q]> + \left<\frac{\partial \hat Q}{\partial t}\left> [/tex]

where the last term is the explicit time dependence of the operator Q.

The Attempt at a Solution



Letting [itex]Q = \vec r \cdot \vec p[/itex]

[tex]\frac{\partial \hat Q}{\partial t} = 0[/tex]

and for stationary states:

[tex]\frac{d}{dt}<Q> = 0[/tex]

so:

[tex]0 = \frac{i}{\hbar}<[\hat H, \hat Q]> = \frac{i}{\hbar}<[T+V, \vec r \cdot \vec p]>[/tex]

[tex]= \frac{i}{\hbar}(<T(\vec r \cdot \vec p)>-<\vec r \cdot \vec p T> + <V(\vec r \cdot \vec p)> - <\vec r \cdot \vec p V>)[/tex]

but

[tex]<\vec r \cdot \vec p V> = <\vec r \cdot (\vec pV)> + <V(\vec r \cdot \vec p)>[/tex]

so

[tex]0 = \frac{i}{\hbar}(<T(\vec r \cdot \vec p)> - <\vec r \cdot \vec p T> - <\vec r \cdot (\vec p V)>)[/tex]
 
Last edited:
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First you show that
$$\begin{align}\frac{d\langle \mathbf{x}\cdot\mathbf{p}\rangle}{dt}=-\frac{i}{\hbar}\sum_{j=1}^3\left(\langle x_j[p_j,H]+[x_j,H]p_j\rangle \right).\end{align}$$
Second you show that $$\begin{align}[p_j,H]=+\frac{\hbar}{i}\frac{\partial V}{\partial x_j}~~\text{and}~~[x_j,H]=-\frac{p_j}{m}.\end{align}$$
Third you substitute equations (2) into (1) and it should pop out.
 

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